Insert all the implicit ( )s and λs into the following abbreviated expressions. Don't just insert them freely; rather, provide the official expression, without any notational shortcuts, that is syntactically identical to the form presented.

In response to your feedback and questions, we refined the explanation of the conventions governing the use of the . shorthand. Thanks!

  1. x x (x x x) x
    (((x x) ((x x) x)) x)
  2. v w (\x y. v x)
    ((v w) (\x (\y (v x))))
  3. (\x y. x) u v
    (((\x (\y x)) u) v)
  4. w (\x y z. x z (y z)) u v
    (((w (\x (\y (\z ((x z) (y z)))))) u) v)

Mark all occurrences of (x y) in the following terms:

  1. (\x y. x y) x y
  2. (\x y. x y) (x y)
  3. \x y. x y (x y)


Find "normal forms" for the following---that is, reduce them until no more reductions are possible. As mentioned in the notes, we'll write λx as \x. If we ever say "reduce" without qualifications, we mean just "beta-reduce" (as opposed to "(beta + eta)-reduce").

  1. (\x \y. y x) z ~~> \y. y z
  2. (\x (x x)) z ~~> z z
  3. (\x (\x x)) z ~~> \x x
  4. (\x (\z x)) z ~~> \y z, be sure to change \z to a different variable so as not to "capture" z
  5. (\x (x (\y y))) (\z (z z)) ~~> \y y
  6. (\x (x x)) (\x (x x)) umm..., reductions will forever be possible, they just don't "do" much
  7. (\x (x x x)) (\x (x x x)) that's just mean


For these questions, and the ones on triples below, we're setting them up so as to encourage you to experiment with Racket and to formulate your answer in Scheme/Racket syntax. But you can answer in Lambda Calculus syntax if you prefer.

Recall our definitions of true and false.

true is defined to be \t f. t
false is defined to be \t f. f

In Racket, these functions can be defined like this:

(define true (lambda (t) (lambda (f) t)))
(define false (lambda (t) (lambda (f) f)))

(Note that they are different from Racket's primitive boolean values #t and #f.)

  1. Define a neg operator that negates true and false.

    Expected behavior:

    (((neg true) 10) 20)

    evaluates to 20, and

    (((neg false) 10) 20)

    evaluates to 10.

    (define neg (lambda (p) ((p false) true)))
  2. Define an or operator.

    (define or (lambda (p) (lambda (q) ((p p) q))))


    (define or (lambda (p) (lambda (q) ((p true) q))))
  3. Define an xor operator. If you haven't seen this term before, here's a truth table:

    true   xor  true   == false
    true   xor  false  == true
    false  xor  true   == true
    false  xor  false  == false

    (define xor (lambda (p) (lambda (q) ((p (neg q)) q))))


Recall our definitions of ordered triples.

the triple (a, b, c) is defined to be \f. f a b c

To extract the first element of a triple t, you write:

t (\fst snd trd. fst)

Here are some definitions in Racket:

(define make-triple  (lambda (fst) (lambda (snd) (lambda (trd) (lambda (f) (((f fst) snd) trd))))))
(define fst_of_three (lambda (fst) (lambda (snd) (lambda (trd) fst))))
(define snd_of_three (lambda (fst) (lambda (snd) (lambda (trd) snd))))

Now we can write:

(define t (((make-triple 10) 20) 30))
(t fst_of_three)  ; will evaluate to 10
(t snd_of_three)  ; will evaluate to 20

If you're puzzled by having the triple to the left and the function that operates on it come second, think about why it's being done this way: the triple is a package that takes a function for operating on its elements as an argument, and returns the result of operating on its elements with that function. In other words, the triple is a higher-order function.

  1. Define the swap12 function that permutes the elements of a triple. Expected behavior:

    (define t (((make-triple 10) 20) 30))
    ((t swap12) fst_of_three) ; evaluates to 20
    ((t swap12) snd_of_three) ; evaluates to 10

    Write out the definition of swap12 in Racket.

    (define swap12 (lambda (x) (lambda (y) (lambda (z)
                     (lambda (f) (((f y) x) z))))))
  2. Define a dup3 function that duplicates its argument to form a triple whose elements are the same. Expected behavior:

      ((dup3 10) fst_of_three) ; evaluates to 10
      ((dup3 10) snd_of_three) ; evaluates to 10

      (define dup3 (lambda (x)
                     (lambda (f) (((f x) x) x))))
  3. Define a dup27 function that makes twenty-seven copies of its argument (and stores them in a data structure of your choice).

    OK, then we will store them in a triply-nested triple:

      (define dup27 (lambda (x) (dup3 (dup3 (dup3 x)))))

Folds and Lists

  1. Using Kapulet syntax, define fold_left.

    # fold_left (f, z) [a, b, c] == f (f (f z a) b) c
      fold_left (f, z) xs = case xs of
                              []       then z;
                              x' & xs' then fold_left (f, f (z, x')) xs'
    in fold_left
  2. Using Kapulet syntax, define filter (problem 7 in last week's homework) in terms of fold_right and other primitive syntax like lambda, &, and []. Don't use letrec! All the letrec-ing that happens should come from the one inside the definition of fold_right.

      filter (p, xs) = fold_right ((lambda (y, ys). if p y then y & ys else ys), []) xs
    in filter
  3. Using Kapulet syntax, define && in terms of fold_right. (To avoid trickiness about the infix syntax, just call it append.) As with problem 22 (the previous problem), don't use letrec!

      xs && ys = fold_right ((&), ys) xs
      # or append (xs, ys) = ...
    in (&&)
  4. Using Kapulet syntax, define head in terms of fold_right. When applied to a non-empty list, it should give us the first element of that list. When applied to an empty list, let's say it should give us 'err. As with problem 22, don't use letrec!

      head xs = fold_right ((lambda (y, _). y), 'err) xs
    in head
  5. We mentioned in the Encoding notes that fold_left (flipped_cons, []) xs would give us the elements of xs but in the reverse order. So that's how we can express reverse in terms of fold_left. How would you express reverse in terms of fold_right? As with problem 22, don't use letrec!

    Here is a boring, inefficient answer

      append (ys, zs) = fold_right ((&), zs) ys;  # aka (&&)
      f (y, prev) = append (prev, [y]);
      reverse xs = fold_right (f, []) xs
    in reverse

    or (same basic idea, just written differently):

      f (y, prev) = fold_right ((&), [y]) prev;
      reverse xs = fold_right (f, []) xs
    in reverse

    Here is an elegant, efficient answer following the hint

    Suppose the list we want to reverse is [10, 20, 30]. Applying fold_right to this will begin by computing f (30, z) for some f and z that we specify. If we made the result of that be something like 30 & blah, or any larger structure that contained something of that form, it's not clear how we could, using just the resources of fold_right, reach down into that structure and replace the blah with some other element, as we'd evidently need to, since after the next step we should get 30 & (20 & blah). What we'd like instead is something like this:

    30 & < >

    Where < > isn't some value but rather a hole. Then with the next step, we want to plug into that hole 20 & < >, which contains its own hole. Getting:

    30 & (20 & < >)

    And so on. That is the key to the solution. The questions you need to answer, to turn this into something executable, are:

    1. What is a hole? How can we implement it?

      A hole is a bound variable. 30 & < > is lambda x. 30 & x.

    2. What should f be, so that the result of the second step, namely f (20, 30 & < >), is 30 & (20 & < >)?

        f (y, prev) = lambda x. prev (y & x)
      in ...
    3. Given that choice of f, what should z be, so that the result of the first step, namely f (30, z) is 30 & < >?

      The identity function: f (30, (lambda y. y)) will reduce to lambda x. (lambda y. y) (30 & x), which will reduce to lambda x. 30 & x.

    4. At the end of the fold_right, we're going to end with something like 30 & (20 & (10 & < >)). But what we want is [30, 20, 10]. How can we turn what we've gotten into what we want?

      Supply it with [] as an argument.

    5. So now put it all together, and explain how to express reverse xs using fold_right and primitive syntax like lambda, &, and []?

        f (y, prev) = lambda x. prev (y & x);
        id match lambda y. y;
        reverse xs = (fold_right (f, id) xs) []
      in reverse

    The ideas here are explored further in Chapter 8 of The Little Schemer. There they first introduce the idea of passing function as arguments to other functions, and having functions be the return values from functions. Then the multirember&co function discussed on pp. 137--140 (and the other ...&co functions discussed in that chapter) are more complex examples of the kind of strategy used here to define reverse. We will be returning to these ideas and considering them more carefully later in the term.


  1. Given that we've agreed to Church's encoding of the numbers:

    0 ≡ \f z. z
    1 ≡ \f z. f z
    2 ≡ \f z. f (f z)
    3 ≡ \f z. f (f (f z))

    How would you express the succ function in the Lambda Calculus?

    let succ = \n. \f z. f (n f z) in ...

    Compare the definition of cons, which has an additional element:

    let cons = \d ds. \f z. f d (ds f z) in ...